思维+multiset ZOJ Monthly, July 2015 - H Twelves Monkeys

 

题目传送门

 1 /*  2  题意：n个时刻点，m次时光穿梭，告诉的起点和终点，q次询问，每次询问t时刻t之前有多少时刻点是可以通过两种不同的路径到达  3  思维：对于当前p时间，从现在到未来穿越到过去的是有效的值，排个序，从大到小询问，那么之前添加的穿越点都是有效的，  4  用multiset保存。比赛时想到了排序，但是无法用线段树实现查询，stl大法好！  5 */  6 #include <cstdio>  7 #include <algorithm>  8 #include <cstring>  9 #include <set> 10 #include <iostream> 11 using namespace std; 12 13 const int MAXN = 1e5 + 10; 14 const int INF = 0x3f3f3f3f; 15 struct Year { 16 int u, v; 17 bool operator < (const Year &r) const { 18 return u > r.u; 19  } 20 }y[MAXN]; 21 struct Query { 22 int p, id; 23 bool operator < (const Query &r) const { 24 return p > r.p; 25  } 26 }q[MAXN]; 27 int ans[MAXN]; 28 int a[3]; 29 int n, m, l; 30 31 int main(void) { //ZOJ Monthly, July 2015 - H Twelves Monkeys 32 //freopen ("H.in", "r", stdin); 33 34 while (scanf ("%d%d%d", &n, &m, &l) == 3) { 35 for (int i=1; i<=m; ++i) { 36 scanf ("%d%d", &y[i].u, &y[i].v); 37  } 38 for (int i=1; i<=l; ++i) { 39 scanf ("%d", &q[i].p); q[i].id = i; 40  } 41 42 sort (y+1, y+1+m); sort (q+1, q+1+l); 43 multiset<int> S; int j = 1; 44 for (int i=1; i<=l; ++i) { 45 while (j <= m && y[j].u >= q[i].p) { 46 S.insert (y[j].v); j++; 47  } 48 multiset<int>::iterator it; int cnt = 0; 49 for (it=S.begin (); it!=S.end (); ++it) { 50 a[++cnt] = *(it); if (cnt == 2) break; 51  } 52 if (cnt == 2 && a[2] <= q[i].p) ans[q[i].id] = q[i].p - a[2]; 53 else ans[q[i].id] = 0; 54  } 55 for (int i=1; i<=l; ++i) 56 printf ("%d\n", ans[i]); 57  } 58 59 return 0; 60 }